Nice problem. Is there an elegant solution? (Find curve for set of chords' midpoints)

[apple2357]

It turns out the resulting quadratics is y=2x^2

This is how i did it but it feels like there might be a neater solution so i'd be interested if anyone has the time/inclination for alternative approaches!

SPOILER BELOW... ( I have missed out some steps for brevity)

Let P be a point on the curve y= x^2-1
So P = (a, a^2-1)
The equation of the line through P passing through the origin is y= (a^2-1)/a x
This line intersects the curve again when ax^2-(a^2-1)x-a=0, call this point Q
Solving this:
x=-1/a and therefore y= (1/a^2)-1
Therfore Coords of Q = (-1/a, 1/a^2-1)
midpoint of PQ = ( (a-1/a)/2 , (a^2+1/a^2-2)/2)
A bit of algebraic manipulation and it follows y= 2x^2

 

[blamocur]

This looks marginally more elegant to me, but this, of course, is "in the eye of the beholder". Symbols [imath]\bar x, \bar y[/imath] are used for the coordinates of the midpoint, and [imath]t[/imath] is the slope of the chord:
[math]y = 2tx \Rightarrow x^2-1 = 2tx \Rightarrow (x-t)^2 = 1+t^2[/math][math]x = t \pm \sqrt{1+t^2} \Rightarrow \bar x = t[/math][math]y = 2tx = 2t^2 \pm 2t\sqrt{1+t^2} \Rightarrow \bar y = 2t^2 = 2x^2[/math]